Problem: You have found the following ages (in years) of 6 bears. Those bears were randomly selected from the 41 bears at your local zoo: $ 31,\enspace 16,\enspace 19,\enspace 13,\enspace 38,\enspace 18$ Based on your sample, what is the average age of the bears? What is the standard deviation? You may round your answers to the nearest tenth.
Answer: Because we only have data for a small sample of the 41 bears, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$ To find the sample mean , add up the values of all $6$ samples and divide by $6$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\overline{x}} = \dfrac{31 + 16 + 19 + 13 + 38 + 18}{{6}} = {22.5\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated {72.25} + {42.25} + {12.25} + {90.25} + {240.25} + {20.25}} {{6 - 1}} $ {s^2} = \dfrac{{477.5}}{{5}} = {95.5\text{ years}^2} $ As you might guess from the notation, the sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$ ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{95.5\text{ years}^2}} = {9.8\text{ years}} $ We can estimate that the average bear at the zoo is 22.5 years old. There is also a standard deviation of 9.8 years.